Giúp mình giai chi tiết với ạ

Giải thích các bước giải:

$\begin{aligned}
& \text{1)} \ \dfrac{1}{y(x-y)} - \dfrac{1}{x(x-y)} = \dfrac{x - y}{xy(x-y)} = \dfrac{1}{xy} \\
& \text{2)} \ \dfrac{x+6}{2(x+3)} + \dfrac{2x+3}{x(x+3)} = \dfrac{x(x+6) + 2(2x+3)}{2x(x+3)} = \dfrac{x^2+10x+6}{2x(x+3)} \\
& \text{3)} \ \dfrac{x}{5(x+1)} - \dfrac{x}{10(x-1)} = \dfrac{2x(x-1) - x(x+1)}{10(x-1)(x+1)} = \dfrac{x^2-3x}{10(x^2-1)} \\
& \text{4)} \ \dfrac{1}{x(y-x)} - \dfrac{1}{y(y-x)} = \dfrac{y - x}{xy(y-x)} = \dfrac{1}{xy} \\
& \text{5)} \ \dfrac{1}{y(y-x)} + \dfrac{1}{x(x-y)} = \dfrac{1}{y(y-x)} - \dfrac{1}{x(y-x)} = \dfrac{x - y}{xy(y-x)} = \dfrac{-(y-x)}{xy(y-x)} = -\dfrac{1}{xy} \\
& \text{6)} \ \dfrac{x}{y(y-x)} - \dfrac{y}{x(y-x)} = \dfrac{x^2 - y^2}{xy(y-x)} = \dfrac{(x-y)(x+y)}{-xy(x-y)} = -\dfrac{x+y}{xy} \\
& \text{7)} \ \dfrac{x+3}{(x-1)(x+1)} - \dfrac{x+1}{x(x-1)} = \dfrac{x(x+3) - (x+1)^2}{x(x-1)(x+1)} = \dfrac{x^2+3x-x^2-2x-1}{x(x^2-1)} = \dfrac{x-1}{x(x-1)(x+1)} = \dfrac{1}{x(x+1)} \\
& \text{8)} \ \dfrac{x+9}{(x-3)(x+3)} - \dfrac{3}{x(x+3)} = \dfrac{x(x+9) - 3(x-3)}{x(x-3)(x+3)} = \dfrac{x^2+6x+9}{x(x^2-9)} = \dfrac{(x+3)^2}{x(x-3)(x+3)} = \dfrac{x+3}{x^2-3x} \\
& \text{9)} \ \dfrac{x-12}{6(x-6)} + \dfrac{6}{x(x-6)} = \dfrac{x(x-12) + 36}{6x(x-6)} = \dfrac{(x-6)^2}{6x(x-6)} = \dfrac{x-6}{6x} \\
& \text{10)} \ \dfrac{3x+5}{x(x-5)} + \dfrac{25-x}{5(5-x)} = \dfrac{3x+5}{x(x-5)} - \dfrac{25-x}{5(x-5)} = \dfrac{5(3x+5) - x(25-x)}{5x(x-5)} = \dfrac{x^2-10x+25}{5x(x-5)} = \dfrac{(x-5)^2}{5x(x-5)} = \dfrac{x-5}{5x} \\
& \text{11)} \ \dfrac{3}{2(x+3)} - \dfrac{x-6}{2x(x+3)} = \dfrac{3x - (x-6)}{2x(x+3)} = \dfrac{2x+6}{2x(x+3)} = \dfrac{2(x+3)}{2x(x+3)} = \dfrac{1}{x} \\
& \text{12)} \ \dfrac{y}{x(2x-y)} + \dfrac{4x}{y(y-2x)} = \dfrac{y}{x(2x-y)} - \dfrac{4x}{y(2x-y)} = \dfrac{y^2 - 4x^2}{xy(2x-y)} = \dfrac{-(2x-y)(2x+y)}{xy(2x-y)} = -\dfrac{2x+y}{xy} \\
& \text{15)} \ \dfrac{7}{x} - \dfrac{x}{x+6} + \dfrac{36}{x(x+6)} = \dfrac{7(x+6) - x^2 + 36}{x(x+6)} = \dfrac{-x^2+7x+78}{x^2+6x} \\
& \text{17)} \ \dfrac{1}{x} + \dfrac{1}{x+5} + \dfrac{x-5}{x(x+5)} = \dfrac{x+5 + x + x-5}{x(x+5)} = \dfrac{3x}{x(x+5)} = \dfrac{3}{x+5} \\
& \text{19)} \ \dfrac{x-4}{x(x-2)} + \dfrac{1}{x-2} - \dfrac{1}{x} = \dfrac{x-4 + x - (x-2)}{x(x-2)} = \dfrac{x-2}{x(x-2)} = \dfrac{1}{x} \\
\end{aligned}$